
Easy basics of vibration diagnostics. Beginner’s guide. Radek Sglunda Adéla Štefková Radim Lang
Open the catalog to page 1- RMS - Root Mean Square
Open the catalog to page 2There‘s nothing to be afraid of If you are faced with the task of implementing and performing vibration diagnostics in your company, then do not panic. This area offers both simple and complex procedures and methods. You will start with the simplest ones and gradually work your way up to more complex ones. At the beginning, you will perform measurements very similar to those made with a voltmeter. Only you will not measure voltage, current or resistance, but vibration values. The basic task of diagnostics is to prevent unexpected failures, because eliminating their consequences entails high costs....
Open the catalog to page 5But what methods can be used to measure the condition of the machine? Vibration diagnostics has an absolutely dominant role in measuring the condition of the machine. It can accurately measure the severity of all basic machine failures. In the following chapters, we will explain everything that is needed for its use. We will not go into theoretical details and mathematical formulas. This is not necessary for the successful implementation of vibration measurements and the subsequent evaluation of the machine condition. In the past, acceptable vibration was tested by putting a coin on a machine....
Open the catalog to page 6intervention as soon as possible. Exceeding the danger limit means the machine must be shut down immediately and repaired. The basic standard is ISO 20816. What if there are no standards? Then other procedures are needed. If we have several identical or similar machines, then we can compare the values between them. If the vibrations on five of six identical machines have a value of 2 and on the sixth the value is 8, then there is clearly something wrong with the sixth machine. Another option is to analyze the trend of the development of vibration values. If the trend is stable over the long time,...
Open the catalog to page 7Time signal Vibrations are the oscillating, repeated movement of mass between two extreme positions. There is nothing unclear in this definition. The important is whether the amplitude of the vibrations (it means the distance between the extreme positions) is acceptable for the machine operation. The velocity of the oscillating movement is also important. A spring and a mass hanging on it is a good example for understanding vibrations and their properties. If we stretch and release the spring, the mass will start to oscillate up and down. The oscillation will decrease due to air resistance and...
Open the catalog to page 8Vibration amplitude The amplitude of the vibration can be easily measured using the 0-P amplitude. We call it the peak value and it is the distance from the mean value of the signal (which corresponds to the rest position of the weight) to the maximum value. In the figure, the 0-P amplitude is equal to the number 5. For completeness, we should mention the P-P (Peak-Peak) value, which is the distance between the maximum and minimum values. For a symmetrical signal shape, the P-P value is twice the 0-P value. But many signals are asymmetrical and this relationship is not true. Signal digitization...
Open the catalog to page 9Sampling frequency First of all, we need to read values at regular (equal) time intervals. For example, the interval between readings will be 1 ms (1ms = 0.001sec). If the signal length is, for example, 1 sec, then we will get 1000 samples, it is a series of 1000 numbers. In the future, we will always call individual readings in a time signal as samples. If we have enough samples, then we can apply various mathematical formulas to them and we can calculate other diverse results. AVG – Average value So far, we can measure the amplitude of time signal as 0-P or P-P values. The amplitude of time...
Open the catalog to page 10For the curious readers, we also present the mathematical formula for a continuous signal waveform, which we will denote as f(x). T In our case, when f(x) = Ι sin(x) Ι then We can imagine the average value also in another way. Let‘s fill in and measure the area under the sine function. Mathematically, this is the calculation of a definite integral, as we mentioned above. Now we will draw a signal whose samples all have the same value. The area under the signal is calculated as the length multiplied by this sample value. It is the area of the rectangle. Our task is to find such a value that the...
Open the catalog to page 11Influence of signal shape on average value We will explain it with the following examples. We have already described the case of a sinusoidal waveform. The average value is equal to the value 0-P multiplied by 0.64. When we calculate the average value from a rectangular waveform, then after converting to an absolute value we see almost the same as after converting to a rectangle. It follows that the average value AVG is almost the same (only slightly smaller) as the value of 0-P. 5 4 3 2 1 0 -1 -2 -3 -4 -5
Open the catalog to page 12If the signal contains only shocks, the situation is the opposite. The area inside the pulses is small, and thus the height of the rectangle for deriving the average AVG value is small. 5 4 3 2 1 0 -1 -2 -3 -4 -5 We have shown that there is no fixed relationship between the average and peak values. RMS value The average value would work well as an indicator of the size of vibrations, but it is not used in practice. A similar value is used, which we call RMS (Root Mean Square). Its advantage is that it corresponds to the energy contained in the signal. Let‘s imagine the level of the force that...
Open the catalog to page 13In the figure we see the signal sin2. The maximum amplitude has increased to 25. This corresponds to 52, when the maximum value was 5 in the original signal. Then we calculate the average value and take the square root of it. If the waveform of the signal corresponds to the sine function, then the RMS value is equal to the peak value 0-P multiplied by 0.71. With a peak value of 5, we get RMS=3.55. This can be expressed again as a constant signal (rectangle) with a height of 3.55 and a width equal to the duration of the signal. I would like to remind you again that the conversion constant 0.71...
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